I got into a discussion with my math teacher. Given the function:
$$ f(x)=a(x-h)^2+k $$
If $h = 3$, $k = 2$ and $a = 1$, then:
$$ f(x)=(x-3)^2+2 $$
We can agree that, when graphing the function, the coordinates of the vertex would be $(3,2)$. The thing is that my teacher wrote in the board that, then:
$$ \text{Vertex} = (h,k) $$
I say it should be:
$$ \text{Vertex}=(-h,k) $$
Which one is syntactically correct? Is the second expression "redundant" or something?
The Vertex as shown here is an actual point and is the point where the quadratic function reaches its minimum value. Given that this is a quadratic function, it has exactly one minimum, so only one representation is correct.
For the function $f(x)=(x-3)^2+2$, using $x=3$ produces $f(3)=(3-3)^2+2=2$ which is the point $(3,2)$. But using $x=-3$ produces $f(-3)=(-3-3)^2+2=38$ which is the point $(-3,38)$ and is clearly not a minimum value for the function and is not equivalent to the point $(3,2)$.
The reason this is the case is due to the fact that in the original formulation, $h$ already has a negative applied to it. In particular, this means that when $x=h$ the value of $a(x-h)^2=0$ which is exactly the $x$-value of the Vertex in all cases. It is applied in this way so that $x$ can take on increasing values if $h$ takes on increasing values, the way that $y$ takes on increasing values if $k$ takes on increasing values.
So the answer is "no, this is not an equivalent way to represent the Vertex."